Is this a good line of best fit?

Theo had a balance of -4 in his savings account. After making a deposit , he has $25 in his account . What is the overall change

ozzi

Theos initial balance in his savings account = -$4
Final balance after making a deposit = $25
So, the overall change to his account = $ 29

6 0

2 years ago

What’s the answer to this question???????

saveliy_v [14]

Answer:

26

Step-by-step explanation:

if YZ bisects VW then VX = XW

if VX = 26 then XW = 26 also

5 0

2 years ago

Find the area of the parallelogram.

Ivenika [448]

Answer:

132 cm

Step-by-step explanation:

22x6=132

7 0

2 years ago

From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi

Lana71 [14]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

$P=\frac{45}{70}= 0.64$

$(1-P)=1-0.64=0.36$

$n= 70$

$P$ ± $z$ $\sqrt{\frac{P(1-P)}{n} }$

where the value $z$ will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of $z$ corresponding to 0.0975 is 1.96

$0.64$ ± $1.96$ $\sqrt{\frac{0.64*0.36}{70} }$

$0.64$ ± $1.96$ $(0.057)$

$0.64$ ± $0.112$

$64%$ % ± $11.2$ %

so the confidence interval is

$64+11.2=75.2$ %

$64-11.2=52.8$ %

$[52.8, 75.2]$

8 0

2 years ago

In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom

coldgirl [10]

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .

Step-by-step explanation:

1) Data given and notation n

n=106 represent the random sample taken

X=48 represent the people who says that they would watch one of the television shows.

$\hat p=\frac{48}{106}=0.453$ estimated proportion of people who says that they would watch one of the television shows.

$p_o=0.5$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 50% of people who says that they would watch one of the television shows.:

Null hypothesis: $p=0.5$

Alternative hypothesis: $p \neq 0.5$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968$

4) Statistical decision

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level is not provided, but we can assume $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .

6 0

3 years ago